Answer:
1257.45 L
Explanation:
We are given;
- Initial volume of Helium gas, V1 as 806 L
- Initial temperature of Helium gas,T1 as 20.9°C
- Initial pressure of Helium gas, P1 as 753 mmHg
- Pressure of Helium at the altitude 6.8 km, P2 as 417 mmHg
- Temperature of Helium gas at the altitude 6.8 Km, T2 as -19.1°C
But, K = °C + 273.15
Therefore, T1 = 294.05 K and T2 = 254.05 K
- We are required to calculate the new volume of the balloon at 6.8 km.
- To determine the new volume we are going to use the combined gas law.
- According to the combined gas law, [tex]\frac{P1V1}{T1}=\frac{P2V2}{T2}[/tex]
Thus, rearranging the formula;
[tex]V2=\frac{P1V1T2}{P2T1}[/tex]
[tex]V2=\frac{(753)(806L)(254.05K)}{(417)(294.05)}[/tex]
[tex]V2=1257.45L[/tex]
Therefore, the volume of the balloon at an altitude of 6.8 km is 1257.45 L
